\(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^5} \, dx\) [400]

Optimal result

Integrand size = 33, antiderivative size = 257 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=-\frac {(B d-A e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e (b d-a e) (d+e x)^4}+\frac {B (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^3}-\frac {3 b B (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^2}+\frac {3 b^2 B (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \] Output:

-1/4*(-A*e+B*d)*(b*x+a)^3*((b*x+a)^2)^(1/2)/e/(-a*e+b*d)/(e*x+d)^4+1/3*B*( 
-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^3-3/2*b*B*(-a*e+b*d)^2*( 
(b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^2+3*b^2*B*(-a*e+b*d)*((b*x+a)^2)^(1/2 
)/e^5/(b*x+a)/(e*x+d)+b^3*B*((b*x+a)^2)^(1/2)*ln(e*x+d)/e^5/(b*x+a)
 

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=-\frac {\sqrt {(a+b x)^2} \left (a^3 e^3 (3 A e+B (d+4 e x))+3 a^2 b e^2 \left (A e (d+4 e x)+B \left (d^2+4 d e x+6 e^2 x^2\right )\right )+3 a b^2 e \left (A e \left (d^2+4 d e x+6 e^2 x^2\right )+3 B \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+b^3 \left (3 A e \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )-B d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )-12 b^3 B (d+e x)^4 \log (d+e x)\right )}{12 e^5 (a+b x) (d+e x)^4} \] Input:

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^5,x]
 

Output:

-1/12*(Sqrt[(a + b*x)^2]*(a^3*e^3*(3*A*e + B*(d + 4*e*x)) + 3*a^2*b*e^2*(A 
*e*(d + 4*e*x) + B*(d^2 + 4*d*e*x + 6*e^2*x^2)) + 3*a*b^2*e*(A*e*(d^2 + 4* 
d*e*x + 6*e^2*x^2) + 3*B*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)) + b^ 
3*(3*A*e*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) - B*d*(25*d^3 + 88*d^ 
2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3)) - 12*b^3*B*(d + e*x)^4*Log[d + e*x])) 
/(e^5*(a + b*x)*(d + e*x)^4)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.62, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {1187, 27, 87, 49, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{(d+e x)^5} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{(d+e x)^5}dx}{b^3 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^5}dx}{a+b x}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {B \int \frac {(a+b x)^3}{(d+e x)^4}dx}{e}-\frac {(a+b x)^4 (B d-A e)}{4 e (d+e x)^4 (b d-a e)}\right )}{a+b x}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {B \int \left (\frac {b^3}{e^3 (d+e x)}-\frac {3 (b d-a e) b^2}{e^3 (d+e x)^2}+\frac {3 (b d-a e)^2 b}{e^3 (d+e x)^3}+\frac {(a e-b d)^3}{e^3 (d+e x)^4}\right )dx}{e}-\frac {(a+b x)^4 (B d-A e)}{4 e (d+e x)^4 (b d-a e)}\right )}{a+b x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {B \left (\frac {3 b^2 (b d-a e)}{e^4 (d+e x)}-\frac {3 b (b d-a e)^2}{2 e^4 (d+e x)^2}+\frac {(b d-a e)^3}{3 e^4 (d+e x)^3}+\frac {b^3 \log (d+e x)}{e^4}\right )}{e}-\frac {(a+b x)^4 (B d-A e)}{4 e (d+e x)^4 (b d-a e)}\right )}{a+b x}\)

Input:

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^5,x]
 

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/4*((B*d - A*e)*(a + b*x)^4)/(e*(b*d - a 
*e)*(d + e*x)^4) + (B*((b*d - a*e)^3/(3*e^4*(d + e*x)^3) - (3*b*(b*d - a*e 
)^2)/(2*e^4*(d + e*x)^2) + (3*b^2*(b*d - a*e))/(e^4*(d + e*x)) + (b^3*Log[ 
d + e*x])/e^4))/e))/(a + b*x)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 1187

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ 
IntPart[p]*(b/2 + c*x)^(2*FracPart[p]))   Int[(d + e*x)^m*(f + g*x)^n*(b/2 
+ c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 
 - 4*a*c, 0] &&  !IntegerQ[p]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.16

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{2} \left (A b e +3 B a e -4 B b d \right ) x^{3}}{e^{2}}-\frac {3 b \left (A a b \,e^{2}+A \,b^{2} d e +B \,e^{2} a^{2}+3 B a b d e -6 B \,b^{2} d^{2}\right ) x^{2}}{2 e^{3}}-\frac {\left (3 A \,a^{2} b \,e^{3}+3 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +B \,e^{3} a^{3}+3 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e -22 B \,b^{3} d^{3}\right ) x}{3 e^{4}}-\frac {3 A \,a^{3} e^{4}+3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +B \,a^{3} d \,e^{3}+3 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e -25 B \,b^{3} d^{4}}{12 e^{5}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{4}}+\frac {b^{3} B \sqrt {\left (b x +a \right )^{2}}\, \ln \left (e x +d \right )}{e^{5} \left (b x +a \right )}\)	\(299\)
default	\(-\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (12 A \,b^{3} e^{4} x^{3}+3 A \,b^{3} d^{3} e +36 B a \,b^{2} e^{4} x^{3}+18 A a \,b^{2} e^{4} x^{2}+18 B \,a^{2} b \,e^{4} x^{2}+12 A \,a^{2} b \,e^{4} x +3 A \,a^{2} b d \,e^{3}+3 A a \,b^{2} d^{2} e^{2}+3 B \,a^{2} b \,d^{2} e^{2}+54 B a \,b^{2} d \,e^{3} x^{2}+9 B a \,b^{2} d^{3} e -12 B \ln \left (e x +d \right ) b^{3} d^{4}-12 B \ln \left (e x +d \right ) b^{3} e^{4} x^{4}+B \,a^{3} d \,e^{3}+4 B \,a^{3} e^{4} x +36 B a \,b^{2} d^{2} e^{2} x +12 A a \,b^{2} d \,e^{3} x +12 B \,a^{2} b d \,e^{3} x -48 B \ln \left (e x +d \right ) b^{3} d^{3} e x -72 B \ln \left (e x +d \right ) b^{3} d^{2} e^{2} x^{2}-48 B \ln \left (e x +d \right ) b^{3} d \,e^{3} x^{3}-48 B \,b^{3} d \,e^{3} x^{3}+18 A \,b^{3} d \,e^{3} x^{2}-108 B \,b^{3} d^{2} e^{2} x^{2}+12 A \,b^{3} d^{2} e^{2} x -88 B \,b^{3} d^{3} e x +3 A \,a^{3} e^{4}-25 B \,b^{3} d^{4}\right )}{12 \left (b x +a \right )^{3} e^{5} \left (e x +d \right )^{4}}\)	\(394\)

Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x,method=_RETURNVERBOSE)
 

Output:

((b*x+a)^2)^(1/2)/(b*x+a)*(-b^2*(A*b*e+3*B*a*e-4*B*b*d)/e^2*x^3-3/2*b*(A*a 
*b*e^2+A*b^2*d*e+B*a^2*e^2+3*B*a*b*d*e-6*B*b^2*d^2)/e^3*x^2-1/3*(3*A*a^2*b 
*e^3+3*A*a*b^2*d*e^2+3*A*b^3*d^2*e+B*a^3*e^3+3*B*a^2*b*d*e^2+9*B*a*b^2*d^2 
*e-22*B*b^3*d^3)/e^4*x-1/12*(3*A*a^3*e^4+3*A*a^2*b*d*e^3+3*A*a*b^2*d^2*e^2 
+3*A*b^3*d^3*e+B*a^3*d*e^3+3*B*a^2*b*d^2*e^2+9*B*a*b^2*d^3*e-25*B*b^3*d^4) 
/e^5)/(e*x+d)^4+b^3*B*((b*x+a)^2)^(1/2)*ln(e*x+d)/e^5/(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=\frac {25 \, B b^{3} d^{4} - 3 \, A a^{3} e^{4} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 12 \, {\left (4 \, B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 18 \, {\left (6 \, B b^{3} d^{2} e^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} - {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 4 \, {\left (22 \, B b^{3} d^{3} e - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} - 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 12 \, {\left (B b^{3} e^{4} x^{4} + 4 \, B b^{3} d e^{3} x^{3} + 6 \, B b^{3} d^{2} e^{2} x^{2} + 4 \, B b^{3} d^{3} e x + B b^{3} d^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x, algorithm="fric 
as")
 

Output:

1/12*(25*B*b^3*d^4 - 3*A*a^3*e^4 - 3*(3*B*a*b^2 + A*b^3)*d^3*e - 3*(B*a^2* 
b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3 + 12*(4*B*b^3*d*e^3 - (3* 
B*a*b^2 + A*b^3)*e^4)*x^3 + 18*(6*B*b^3*d^2*e^2 - (3*B*a*b^2 + A*b^3)*d*e^ 
3 - (B*a^2*b + A*a*b^2)*e^4)*x^2 + 4*(22*B*b^3*d^3*e - 3*(3*B*a*b^2 + A*b^ 
3)*d^2*e^2 - 3*(B*a^2*b + A*a*b^2)*d*e^3 - (B*a^3 + 3*A*a^2*b)*e^4)*x + 12 
*(B*b^3*e^4*x^4 + 4*B*b^3*d*e^3*x^3 + 6*B*b^3*d^2*e^2*x^2 + 4*B*b^3*d^3*e* 
x + B*b^3*d^4)*log(e*x + d))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^ 
3*e^6*x + d^4*e^5)
 

Sympy [F]

\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{5}}\, dx \] Input:

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**5,x)
 

Output:

Integral((A + B*x)*((a + b*x)**2)**(3/2)/(d + e*x)**5, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x, algorithm="maxi 
ma")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (196) = 392\).

Time = 0.22 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.68 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=\frac {B b^{3} \log \left ({\left | e x + d \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{e^{5}} + \frac {12 \, {\left (4 \, B b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 18 \, {\left (6 \, B b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{2} b e^{3} \mathrm {sgn}\left (b x + a\right ) - A a b^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 4 \, {\left (22 \, B b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, B a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x + \frac {25 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 3 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )}{e}}{12 \, {\left (e x + d\right )}^{4} e^{4}} \] Input:

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x, algorithm="giac 
")
 

Output:

B*b^3*log(abs(e*x + d))*sgn(b*x + a)/e^5 + 1/12*(12*(4*B*b^3*d*e^2*sgn(b*x 
 + a) - 3*B*a*b^2*e^3*sgn(b*x + a) - A*b^3*e^3*sgn(b*x + a))*x^3 + 18*(6*B 
*b^3*d^2*e*sgn(b*x + a) - 3*B*a*b^2*d*e^2*sgn(b*x + a) - A*b^3*d*e^2*sgn(b 
*x + a) - B*a^2*b*e^3*sgn(b*x + a) - A*a*b^2*e^3*sgn(b*x + a))*x^2 + 4*(22 
*B*b^3*d^3*sgn(b*x + a) - 9*B*a*b^2*d^2*e*sgn(b*x + a) - 3*A*b^3*d^2*e*sgn 
(b*x + a) - 3*B*a^2*b*d*e^2*sgn(b*x + a) - 3*A*a*b^2*d*e^2*sgn(b*x + a) - 
B*a^3*e^3*sgn(b*x + a) - 3*A*a^2*b*e^3*sgn(b*x + a))*x + (25*B*b^3*d^4*sgn 
(b*x + a) - 9*B*a*b^2*d^3*e*sgn(b*x + a) - 3*A*b^3*d^3*e*sgn(b*x + a) - 3* 
B*a^2*b*d^2*e^2*sgn(b*x + a) - 3*A*a*b^2*d^2*e^2*sgn(b*x + a) - B*a^3*d*e^ 
3*sgn(b*x + a) - 3*A*a^2*b*d*e^3*sgn(b*x + a) - 3*A*a^3*e^4*sgn(b*x + a))/ 
e)/((e*x + d)^4*e^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^5} \,d x \] Input:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^5,x)
 

Output:

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^5, x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^5} \, dx=\frac {12 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{5}+48 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{4} e x +72 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{3} e^{2} x^{2}+48 \,\mathrm {log}\left (e x +d \right ) b^{4} d^{2} e^{3} x^{3}+12 \,\mathrm {log}\left (e x +d \right ) b^{4} d \,e^{4} x^{4}-3 a^{4} d \,e^{4}-4 a^{3} b \,d^{2} e^{3}-16 a^{3} b d \,e^{4} x -6 a^{2} b^{2} d^{3} e^{2}-24 a^{2} b^{2} d^{2} e^{3} x -36 a^{2} b^{2} d \,e^{4} x^{2}+12 a \,b^{3} e^{5} x^{4}+13 b^{4} d^{5}+40 b^{4} d^{4} e x +36 b^{4} d^{3} e^{2} x^{2}-12 b^{4} d \,e^{4} x^{4}}{12 d \,e^{5} \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )} \] Input:

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x)
 

Output:

(12*log(d + e*x)*b**4*d**5 + 48*log(d + e*x)*b**4*d**4*e*x + 72*log(d + e* 
x)*b**4*d**3*e**2*x**2 + 48*log(d + e*x)*b**4*d**2*e**3*x**3 + 12*log(d + 
e*x)*b**4*d*e**4*x**4 - 3*a**4*d*e**4 - 4*a**3*b*d**2*e**3 - 16*a**3*b*d*e 
**4*x - 6*a**2*b**2*d**3*e**2 - 24*a**2*b**2*d**2*e**3*x - 36*a**2*b**2*d* 
e**4*x**2 + 12*a*b**3*e**5*x**4 + 13*b**4*d**5 + 40*b**4*d**4*e*x + 36*b** 
4*d**3*e**2*x**2 - 12*b**4*d*e**4*x**4)/(12*d*e**5*(d**4 + 4*d**3*e*x + 6* 
d**2*e**2*x**2 + 4*d*e**3*x**3 + e**4*x**4))